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Current Question (ID: 20336)

Question:

$\text{Two identical wires } \text{A} \text{ and } \text{B}, \text{ each of length } 'l', \text{ carry the same current } I. \text{ Wire } \text{A} \text{ is bent into a circle of radius } R \text{ and wire } \text{B} \text{ is bent to form a square of side } 'a'. \text{ If } B_A \text{ and } B_B \text{ are the values of the magnetic field at the centres of the circle and square respectively, then the ratio } \frac{B_A}{B_B} \text{ is?}$

Options:

1. $\frac{\pi^2}{8}$
2. $\frac{\pi^2}{16\sqrt{2}}$
3. $\frac{\pi^2}{16}$
4. $\frac{\pi^2}{8\sqrt{2}}$

Solution:

$\text{Hint: } B = \frac{\mu_0 I}{2R} \text{ for the ring.}$ $B_1 = \frac{\mu_0 I}{2R} = \frac{\mu_0 I}{2 \times \frac{l}{2\pi}} = \frac{\mu_0 \pi I}{l}$ $B_2 = \frac{\mu_0 I}{4\pi a} \left[ \sin \alpha + \sin \beta \right] \times 4 = \frac{\mu_0 I}{4\pi \frac{a}{2}} \left[ \sin \right] = 2\sqrt{2} \frac{\mu_0 I}{\pi a} = 2\sqrt{2} \frac{\mu_0 I}{\frac{\pi l}{4}} = 8\sqrt{2} \frac{\mu_0 I}{\pi l}$ $\frac{B_1}{B_2} = \frac{\frac{\mu_0 \pi I}{l}}{8\sqrt{2} \frac{\mu_0 I}{\pi l}} = \frac{\pi^2}{8\sqrt{2}}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}